print barcode image c# Indeterminate Forms in Software

Creator QR Code 2d barcode in Software Indeterminate Forms

CHAPTER 5 Indeterminate Forms
QR Code ISO/IEC18004 Creator In None
Using Barcode creation for Software Control to generate, create Quick Response Code image in Software applications.
QR Code JIS X 0510 Recognizer In None
Using Barcode reader for Software Control to read, scan read, scan image in Software applications.
converges Since the rst integral diverges, we conclude that the original integral diverges as well You Try It: Calculate
Paint QR In C#
Using Barcode printer for .NET framework Control to generate, create QR Code image in VS .NET applications.
QR-Code Printer In Visual Studio .NET
Using Barcode generator for ASP.NET Control to generate, create QR-Code image in ASP.NET applications.
3 1/3 dx 2 (2x)
QR Code Drawer In VS .NET
Using Barcode creator for .NET Control to generate, create QR Code JIS X 0510 image in VS .NET applications.
Encode QR Code JIS X 0510 In Visual Basic .NET
Using Barcode generation for .NET framework Control to generate, create QR image in .NET framework applications.
as an improper integral
Code 128C Creation In None
Using Barcode generation for Software Control to generate, create Code128 image in Software applications.
Encoding Bar Code In None
Using Barcode generation for Software Control to generate, create barcode image in Software applications.
AN APPLICATION TO AREA
UPC-A Printer In None
Using Barcode creation for Software Control to generate, create UPC Code image in Software applications.
EAN128 Printer In None
Using Barcode maker for Software Control to generate, create UCC-128 image in Software applications.
Suppose that f is a non-negative, continuous function on the interval (a, b] which is unbounded as x a + Look at Fig 55 Let us consider the area under the graph of f and above the x-axis over the interval (a, b] The area of the part of the region over the interval [a + , b], > 0, is
Data Matrix 2d Barcode Encoder In None
Using Barcode maker for Software Control to generate, create Data Matrix image in Software applications.
EAN-13 Supplement 5 Creator In None
Using Barcode printer for Software Control to generate, create EAN-13 image in Software applications.
f (x) dx
Identcode Generation In None
Using Barcode generator for Software Control to generate, create Identcode image in Software applications.
Drawing UPC-A In Java
Using Barcode printer for Android Control to generate, create Universal Product Code version A image in Android applications.
Fig 55
Paint Barcode In None
Using Barcode encoder for Online Control to generate, create barcode image in Online applications.
GTIN - 128 Printer In Visual Studio .NET
Using Barcode encoder for Visual Studio .NET Control to generate, create GS1-128 image in VS .NET applications.
Therefore it is natural to consider the area of the entire region, over the interval (a, b], to be
Make USS-128 In Java
Using Barcode maker for Java Control to generate, create GTIN - 128 image in Java applications.
Creating Bar Code In .NET
Using Barcode printer for .NET framework Control to generate, create barcode image in Visual Studio .NET applications.
lim This is just the improper integral
Reading EAN / UCC - 13 In VB.NET
Using Barcode reader for VS .NET Control to read, scan read, scan image in VS .NET applications.
Make Bar Code In Java
Using Barcode generator for Java Control to generate, create bar code image in Java applications.
0+ a+
f (x) dx
Area =
f (x) dx
EXAMPLE 520
Indeterminate Forms
Calculate the area above the x-axis and under the curve y= 1 x ln4/3 x , 0 < x 1/2
SOLUTION According to the preceding discussion, this area is equal to the value of the improper integral
1/2 0
1 x ln
dx = lim
1 x ln4/3 x
For clarity we let (x) = ln x, (x) = 1/x Then the (inde nite) integral becomes (x) 3 dx = 1/3 4/3 (x) (x) Thus
1 x ln4/3 x
dx = lim
3 ln1/3 x
= lim
3 3 1/3 [ ln 2] [ln ]1/3
Now as 0 then ln hence 1/[ln ]1/3 0 We conclude that our improper integral converges and the area under the curve and above the x-axis equals 3/[ln 2]1/3
More on Improper Integrals
541 INTRODUCTION
Suppose that we want to calculate the integral of a continuous function f (x) over an unbounded interval of the form [A, + ) or ( , B] The theory of the integral that we learned earlier does not cover this situation, and some new concepts are needed We treat improper integrals on in nite intervals in this section, and give some applications at the end
CHAPTER 5 Indeterminate Forms 542 THE INTEGRAL ON AN INFINITE INTERVAL
Let f be a continuous function whose domain contains an interval of the form [A, + ) The value of the improper integral f (x) dx
is de ned to be
N N + A
f (x) dx
Similarly, we have: Let g be a continuous function whose domain contains an interval of the form ( , B] The value of the improper integral
B
g(x) dx
is de ned to be
B M M
f (x) dx
EXAMPLE 521
Calculate the improper integral
x 3 dx
SOLUTION We do this problem by evaluating the limit
N N + 1
x 3 dx = =
N +
(1/2)x 2
N +
lim (1/2) N 2 1 2
1 = 2 We conclude that the integral converges and has value 1/2 EXAMPLE 522
Evaluate the improper integral
32
x 1/5 dx
Indeterminate Forms
SOLUTION We do this problem by evaluating the limit
32 M M
x 1/5 dx =
5 4/5 32 x M 4 M 5 ( 32)4/5 M 4/5 = lim M 4 5 16 M 4/5 = lim M 4 lim
This limit equals Therefore the integral diverges You Try It: Evaluate
3 dx 1 (1 + x)
Sometimes we have occasion to evaluate a doubly in nite integral We do so by breaking the integral up into two separate improper integrals, each of which can be evaluated with just one limit EXAMPLE 523
Evaluate the improper integral
2 1 + x
SOLUTION The interval of integration is ( , + ) To evaluate this integral, we break the interval up into two pieces: ( , + ) = ( , 0] [0, + ) (The choice of zero as a place to break the interval is not important; any other point would do in this example) Thus we will evaluate separately the integrals + 0 1 1 dx and dx 2 2 1+x 0 1 + x For the rst one we consider the limit
N N + 0
1 dx = lim Tan 1 x N + 1 + x2 = =
N +
Tan 1 N Tan 1 0
2
CHAPTER 5 Indeterminate Forms
The second integral is evaluated similarly:
0 M M
1 dx = lim Tan 1 x M 1 + x2 = =
M
0 Tan 1 M
2 Since each of the integrals on the half line is convergent, we conclude that the original improper integral over the entire real line is convergent and that its value is + = 2 2 You Try It: Discuss
1 dx 1 (1 + x)
SOME APPLICATIONS
Now we may use improper integrals over in nite intervals to calculate area EXAMPLE 524
Calculate the area under the curve y = 1/[x (ln x)4 ] and above the x-axis, 2 x <
SOLUTION The area is given by the improper integral
+ 2
1 dx = lim N + x (ln x)4
1 dx x (ln x)4
For clarity, we let (x) = ln x, (x) = 1/x Thus the (inde nite) integral becomes 1/3 (x) dx = 3 4 (x) (x) Thus
N N + 2
1 1/3 N dx = lim 3 N + x (ln x)4 ln x 2 1/3 1/3 3 = lim 3 N + ln N ln 2 1/3 = 3 ln 2
Thus the area under the curve and above the x-axis is 1/(3 ln3 2)
EXAMPLE 525
Indeterminate Forms
Because of in ation, the value of a dollar decreases as time goes on Indeed, this decrease in the value of money is directly related to the continuous compounding of interest For if one dollar today is invested at 6% continuously compounded interest for ten years then that dollar will have grown to e006 10 = $182 (see Section 65 for more detail on this matter) This means that a dollar in the currency of ten years from now corresponds to only e 006 10 = $055 in today s currency Now suppose that a trust is established in your name which pays 2t + 50 dollars per year for every year in perpetuity, where t is time measured in years (here the present corresponds to time t = 0) Assume a constant interest rate of 6%, and that all interest is re-invested What is the total value, in today s dollars, of all the money that will ever be earned by your trust account
SOLUTION Over a short time increment [tj 1 , tj ], the value in today s currency of the money earned is about (2tj + 50) e 006 tj The corresponding sum over time increments is (2tj + 50) e 006 tj tj
Copyright © OnBarcode.com . All rights reserved.